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Frame Element
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Outlines Review
FE Formulation of Frame Structure Element
Example
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Objectives Understand what frame element is and its degree of
freedoms
To formulate the stiffness matrix of frame element To solve a frame problem using the stiffness-
displacement-load matrices approach
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Beam element - review1000 lb/ft
10 ft2.5 ft
1 2 3
2.5 ft
500 lb
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ReviewElement Type DOF K matrix Remark
Spring/ bar U1
Truss U1, U2
Beam U1, θ1
Frame U1, U2, θ1
k k
k k K
22
22
22
22
mlmmlm
lml lml
mlmmlm
lml lml
K
22
22
3
4626
612612
2646
612612
L L L L
L L
L L L L
L L
L EI K
L
EAk
sin
cos
m
l
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Definition: So…what ARE frames??
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Definition: Structural members that may be rigidly connected
with welded joints or bolted joints
Combination of bar (truss) and beam
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X
Y i
j
θ
3 DOF at each node: lateral andlongitudinal displacements, androtation
Combination of bar and beam
Superimpose to get the stiffness matrix
Frame Element
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Revisit: Bar and Beam Elements
22
22
3
4626
612612
2646
612612
L L L L
L L
L L L L
L L
L
EI K
k k
k k K
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Truss Element
Stiffness matrix for members under axial loading
3
2
1
3
2
1
000000000000
0000
000000
000000
0000
j
j
j
i
i
i
L AE
L AE
L AE
L AE
axial
uu
u
u
u
u
K
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Beam Element
Stiffness matrix for lateral displacements and rotations
3
2
1
3
2
1
22
22
3
46026061206120
000000
260460
61206120
000000
j
j
j
i
i
i
xy
uu
u
u
u
u
L L L L L L
L L L L
L L
L
EI K
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Stiffness matrix for Frame
+
=
22
22
3
460260
61206120
000000
260460
61206120000000
L L L L
L L
L L L L
L L
L
EI
000000
000000
0000
000000
0000000000
L AE
L AE
L
AE
L
AE
3
2
1
3
2
1
46
26
612612
2646
612612
22
2323
22
2323
00
00
0000
00
00
0000
j
j
j
i
i
i
L EI L EI L EI L
EI
L
EI
L
EI
L
EI
L
EI
L AE
L AE
L EI
L
EI L EI
L
EI
L EI
L EI
L EI
L EI
L AE
L AE
u
u
u
u
u
u
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Local DOF related toGlobal DOF through the
transformation matrix
In Matrix form: {U}=[T]{u}
Trans form ation Matr ix
X
Y
i
j
θ
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cossin00
sincos00
00cossin
00sincos
T
100000
0cossin000
0sincos000
000100
0000cossin
0000sincos
T
In Matrix form: {U}=[T]{u}{F}=[T]{f}
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Stiffness matrix
T eTK TK thatconcludewe
UTk TF
FUTk T
FTTUTk T
FTUTk
f TFanduTU f uk
1
1
11
11
,
From strain energy equation, we can show that;
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Stiffness matrix
100000
0cosθsinθ000
0sinθcosθ000
000100
0000cosθsinθ
0000sinθcosθ
00
00
0000
00
00
0000
100000
0cosθsinθ-000
0sinθcosθ000
000100
0000cosθsinθ-
0000sinθcosθ
K
L
4EI
L
6EI
L
2EI
L
6EI
L
6EI
L
12EI
L
6EI
L
12EI
L
AE
L
AE
L
2EI
L
6EI
L
4EI
L
6EI
L
6EI
L
12EI
L
6EI
L
12EI
L
AE
L
AE
T
e
22
2323
22
2323
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Example 4.5
10 ft
9 ft
800 lb/ft
E=30 x 106 psi
A=7.65 in2
I=204 in4
Determine the deformation of the frame under the
given distributed load.
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Element 1
(1) 21
u22
u21
u23
u12
u11u13
Y
X
L
4EI
L
6EI
L
2EI
L
6EI
L
6EI
L
12EI
L
6EI
L
12EI
L
AE
L
AE
L
2EI
L
6EI
L
4EI
L
6EI
L
6EI
L
12EI
L
6EI
L
12EI
L
AE
L
AE
22
2323
22
2323
00
00
0000
00
00
0000
e
xy K
E =30 x 106 lb/in2
A= 7.65 in2
I = 204 in4
L = 10 ft
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Stiffness matrix with respect to
local C.S
23
22
21
13
12
11
31
2040002550010200025500
255042.50255042.50
001912.5001912.5
1020002550020400025500
25505.420255042.50
005.1912001912.5
10
u
u
u
u
u
u
K xy
Since the local and the global C.S are aligned in the same direction,
[K]1 same as global stiffness matrix.
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Element 2
(2)
3 u32
u31
u33
u22
u21
u23
Y
X
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Local stiffness matrix for element 2
33
32
31
23
22
21
32
2266663148.15-01133333148.150
3148.1558.3-03148.1558.3-0
002125002125
1133333148.1502266663148.150
3148.153.5803148.1558.30
002125-002125
10
u
u
u
uu
u
K xy
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Transformation matrix[T] when Ɵ = 270o
100000
0270cos270sin000
0270sin270cos000
000100
0000270cos270sin
0000270sin270cos
100000
0cossin000
0sincos000
000100
0000cossin
0000sincos
T
100000
001000
010000
000100
000001
000010
T
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The stiffness matrix for element 2
becomes,
100000
001000
010000000100
000001
000010
2266663148.15-01133333148.150
3148.1558.3-03148.1558.3-0002125002125
1133333148.1502266663148.150
3148.153.5803148.1558.30
002125-002125
10
100000
001000010000
000100
000001
000010
32
T
xy K
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Perform matrix operation:
33
32
31
23
22
21
32
22666603148.15-11333303148.15
02125002125-0
3148.1503.583148.15058.3-
11333303148.1522666603148.15
021250021250
3148.1503.583148.15058.3
10
u
u
u
u
u
u
K xy
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Assemble the elements
226666015.3148113333015.3148000
021250021250000
15.314803.5815.314803.58000
113333015.3148226666204000255015.3148010200025500
025500255021255.42025505.420
15.314803.5815.3148003.585.1912005.1912
0001020002550020400025500
00025505.42025505.420
000005.1912005.1912
G K
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Nodal loads
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Solve for unknowns
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Recap Frame structure element has 3 DOFs
Stiffness matrix for frame element is
[K]=[T]T
[k][T]
100000
0cossin000
0sincos000
000100
0000cossin
0000sincos
T
L4EI
L
6EI
L2EI
L
6EI
L
6EI
L
12EI
L
6EI
L
12EI
LAE
LAE
L
2EI
L
6EI
L
4EI
L
6EI
L
6EI
L
12EI
L
6EI
L
12EI
L
AE
L
AE
22
2323
22
2323
00
00
0000
00
00
0000
e
xy K